March 15th, 2009
NG Big Idea
Many important practical and mathematical applications involve comparing quantities of one kind or another; it is important to know which method to use and how we should use them.Essential Question: What methods are there for comparing things? Problem 4.2
When Madeline an Luis compared the fuel economy of their new cars, they found these rates:
Madeline's car went 580 miles with 19 gallons of gasoline.
Luis's car went 452 miles with 15.5 gallons of gasoline.
Use this information to answer the following questions.
A. For each car, find a unit rate describing the mileage. Which car got better gas mileage? In other words, which car went more miles per gallons of gas?
Madeline's car goes 30.5 miles per gallon of gas (580 miles/19 gallons of gas). Luis's car goes 29.2 miles per gallon of gas (452 miles/15.5 gallons of gas). Madeline's car has got better gas mileage.
B. Complete a table, showing the fuel used and the miles covered by each car based on the unit rates you found in part A.
Gallons of gas
0
1
2
3
4
5
6
7
8
Miles in Madeline's car
0
30.5
61
91.5
122
152.5
183
213.5
244
Miles in Luis's car
0
29.2
58.4
87.6
116.8
146
175.2
204.4
233.6
C. Look at the patterns in your table. For each car, write an equation for a rule you can use to predict the miles driven (m) from the gallons of gas used (g).
Madeline's car: 30.5 m/g x g = m
Luis' car: 29.2 m/g x g = m
D. Use the rules you wrote in part C to find the number of miles each car could cover of it used 9.5, 15.5, 19, 23.8, 100, 125, and 150 gallons of gasoline.
Madeline's car:
30.5 m/g x 9.5 g = 289.75 m.
30.5 m/g x 15.5 g = 472.75 m.
30.5 m/g x 19 g = 17,674.75 m.
30.5 m/g x 23.8 g = 725.9 m.
30.5 m/g x 100 g = 3,050 m.
30.5 m/g x 125 g = 3,812.5 m.
30.5 m/g x 150 g = 4,575 m.
Luis's car:
29.2 m/g x 9.5 g = 277.4 m.
29.2 m/g x 15.5 g = 452.6 m.
29.2 m/g x 19 g = 554.8 m.
29.2 m/g x 23.8 g = 694.96 m.
29.2 m/g x 100 g = 2,920 m.
29.2 m/g x 125 g = 3,650 m.
29.2 m/g x 150 g = 4,380 m.
Problem 4.2 Follow Up
1. Use you data from B or D to sketch graphs of the (gallons, miles) data for each car.
2. How are your two graphs alike? How are they different?
3. What do you think makes the two graphs different?
NG
Big Idea
Many important practical and mathematical applications involve comparing quantities of one kind or another; it is important to know which method to use and how we should use them. Essential Question:
What methods are there for comparing things?
Problem 4.2
When Madeline an Luis compared the fuel economy of their new cars, they found these rates:
Madeline's car went 580 miles with 19 gallons of gasoline.
Luis's car went 452 miles with 15.5 gallons of gasoline.
Use this information to answer the following questions.
A. For each car, find a unit rate describing the mileage. Which car got better gas mileage? In other words, which car went more miles per gallons of gas?
Madeline's car goes 30.5 miles per gallon of gas (580 miles/19 gallons of gas). Luis's car goes 29.2 miles per gallon of gas (452 miles/15.5 gallons of gas). Madeline's car has got better gas mileage.
B. Complete a table, showing the fuel used and the miles covered by each car based on the unit rates you found in part A.
Madeline's car: 30.5 m/g x g = m
Luis' car: 29.2 m/g x g = m
D. Use the rules you wrote in part C to find the number of miles each car could cover of it used 9.5, 15.5, 19, 23.8, 100, 125, and 150 gallons of gasoline.
Madeline's car:
30.5 m/g x 9.5 g = 289.75 m.
30.5 m/g x 15.5 g = 472.75 m.
30.5 m/g x 19 g = 17,674.75 m.
30.5 m/g x 23.8 g = 725.9 m.
30.5 m/g x 100 g = 3,050 m.
30.5 m/g x 125 g = 3,812.5 m.
30.5 m/g x 150 g = 4,575 m.
Luis's car:
29.2 m/g x 9.5 g = 277.4 m.
29.2 m/g x 15.5 g = 452.6 m.
29.2 m/g x 19 g = 554.8 m.
29.2 m/g x 23.8 g = 694.96 m.
29.2 m/g x 100 g = 2,920 m.
29.2 m/g x 125 g = 3,650 m.
29.2 m/g x 150 g = 4,380 m.
Problem 4.2 Follow Up
1. Use you data from B or D to sketch graphs of the (gallons, miles) data for each car.
2. How are your two graphs alike? How are they different?
3. What do you think makes the two graphs different?